You have found the following ages (in years) of all 5 bears at your local zoo: $ 6,\enspace 4,\enspace 25,\enspace 35,\enspace 33$ What is the average age of the bears at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Solution: Because we have data for all 5 bears at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $5$ ages and divide by $5$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\mu} = \dfrac{6 + 4 + 25 + 35 + 33}{{5}} = {20.6\text{ years old}} $ Find the squared deviations from the mean for each bear. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $6$ years $-14.6$ years $213.16$ years $^2$ $4$ years $-16.6$ years $275.56$ years $^2$ $25$ years $4.4$ years $19.36$ years $^2$ $35$ years $14.4$ years $207.36$ years $^2$ $33$ years $12.4$ years $153.76$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{213.16} + {275.56} + {19.36} + {207.36} + {153.76}} {{5}} $ $ {\sigma^2} = \dfrac{{869.2}}{{5}} = {173.84\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{173.84\text{ years}^2}} = {13.2\text{ years}} $ The average bear at the zoo is 20.6 years old. There is a standard deviation of 13.2 years.